## PHY2403H Quantum Field Theory. Lecture 14: Time evolution, Hamiltonian perturbation, ground state. Taught by Prof. Erich Poppitz

### DISCLAIMER: Very rough notes from class, with some additional side notes.

These are notes for the UofT course PHY2403H, Quantum Field Theory, taught by Prof. Erich Poppitz, fall 2018.

## Review

Given a field $$\phi(t_0, \Bx)$$, satisfying the commutation relations
\label{eqn:qftLecture14:20}
\antisymmetric{\pi(t_0, \Bx)}{\phi(t_0, \By)} = -i \delta(\Bx – \By)

we introduced an interaction picture field given by
\label{eqn:qftLecture14:40}
\phi_I(t, x) = e^{i H_0(t- t_0)} \phi(t_0, \Bx) e^{-iH_0(t – t_0)}

related to the Heisenberg picture representation by
\label{eqn:qftLecture14:60}
\phi_H(t, x)
= e^{i H(t- t_0)} \phi(t_0, \Bx) e^{-iH(t – t_0)}
= U^\dagger(t, t_0) \phi_I(t, \Bx) U(t, t_0),

where $$U(t, t_0)$$ is the time evolution operator.
\label{eqn:qftLecture14:80}
U(t, t_0) =
e^{i H_0(t – t_0)}
e^{-i H(t – t_0)}

We argued that
\label{eqn:qftLecture14:100}
i \PD{t}{} U(t, t_0) = H_{\text{I,int}}(t) U(t, t_0)

We found the glorious expression
\label{eqn:qftLecture14:120}
\boxed{
\begin{aligned}
U(t, t_0)
&= T \exp{\lr{ -i \int_{t_0}^t H_{\text{I,int}}(t’) dt’}} \\
&=
\sum_{n = 0}^\infty \frac{(-i)^n}{n!} \int_{t_0}^t dt_1 dt_2 \cdots dt_n T\lr{ H_{\text{I,int}}(t_1) H_{\text{I,int}}(t_2) \cdots H_{\text{I,int}}(t_n) }
\end{aligned}
}

However, what we are really after is
\label{eqn:qftLecture14:140}
\bra{\Omega} T(\phi(x_1) \cdots \phi(x_n)) \ket{\Omega}

Such a product has many labels and names, and we’ll describe it as “vacuum expectation values of time-ordered products of arbitrary #’s of local Heisenberg operators”.

## Perturbation

Following section 4.2, [1].

\label{eqn:qftLecture14:160}
\begin{aligned}
H &= \text{exact Hamiltonian} = H_0 + H_{\text{int}}
\\
H_0 &= \text{free Hamiltonian.
}
\end{aligned}

We know all about $$H_0$$ and assume that it has a lowest (ground state) $$\ket{0}$$, the “vacuum” state of $$H_0$$.

$$H$$ has eigenstates, in particular $$H$$ is assumed to have a unique ground state $$\ket{\Omega}$$ satisfying
\label{eqn:qftLecture14:180}
H \ket{\Omega} = \ket{\Omega} E_0,

and has states $$\ket{n}$$, representing excited (non-vacuum states with energies > $$E_0$$).
These states are assumed to be a complete basis
\label{eqn:qftLecture14:200}
\mathbf{1} = \ket{\Omega}\bra{\Omega} + \sum_n \ket{n}\bra{n} + \int dn \ket{n}\bra{n}.

The latter terms may be written with a superimposed sum-integral notation as
\label{eqn:qftLecture14:440}
\sum_n + \int dn
=
{\int\kern-1em\sum}_n,

so the identity operator takes the more compact form
\label{eqn:qftLecture14:460}
\mathbf{1} = \ket{\Omega}\bra{\Omega} + {\int\kern-1em\sum}_n \ket{n}\bra{n}.

For some time $$T$$ we have
\label{eqn:qftLecture14:220}
e^{-i H T} \ket{0} = e^{-i H T}
\lr{
\ket{\Omega}\braket{\Omega}{0} + {\int\kern-1em\sum}_n \ket{n}\braket{n}{0}
}.

We now wish to argue that the $${\int\kern-1em\sum}_n$$ term can be ignored.

### Argument 1:

This is something of a fast one, but one can consider a formal transformation $$T \rightarrow T(1 – i \epsilon)$$, where $$\epsilon \rightarrow 0^+$$, and consider very large $$T$$. This gives
\label{eqn:qftLecture14:240}
\begin{aligned}
\lim_{T \rightarrow \infty, \epsilon \rightarrow 0^+}
e^{-i H T(1 – i \epsilon)} \ket{0}
&=
\lim_{T \rightarrow \infty, \epsilon \rightarrow 0^+}
e^{-i H T(1 – i \epsilon)}
\lr{
\ket{\Omega}\braket{\Omega}{0} + {\int\kern-1em\sum}_n \ket{n}\braket{n}{0}
} \\
&=
\lim_{T \rightarrow \infty, \epsilon \rightarrow 0^+}
e^{-i E_0 T – E_0 \epsilon T}
\ket{\Omega}\braket{\Omega}{0} + {\int\kern-1em\sum}_n e^{-i E_n T – \epsilon E_n T} \ket{n}\braket{n}{0} \\
&=
\lim_{T \rightarrow \infty, \epsilon \rightarrow 0^+}
e^{-i E_0 T – E_0 \epsilon T}
\lr{
\ket{\Omega}\braket{\Omega}{0} + {\int\kern-1em\sum}_n e^{-i (E_n -E_0) T – \epsilon T (E_n – E_0)} \ket{n}\braket{n}{0}
}
\end{aligned}

The limits are evaluated by first taking $$T$$ to infinity, then only after that take $$\epsilon \rightarrow 0^+$$. Doing this, the sum is dominated by the ground state contribution, since each excited state also has a $$e^{-\epsilon T(E_n – E_0)}$$ suppression factor (in addition to the leading suppression factor).

### Argument 2:

With the hand waving required for the argument above, it’s worth pointing other (less formal) ways to arrive at the same result. We can write
\label{eqn:qftLecture14:260}
sectionumInt \ket{n}\bra{n} \rightarrow
\sum_k \int \frac{d^3 p}{(2 \pi)^3} \ket{\Bp, k}\bra{\Bp, k}

where $$k$$ is some unknown quantity that we are summing over.
If we have
\label{eqn:qftLecture14:280}
H \ket{\Bp, k} = E_{\Bp, k} \ket{\Bp, k},

then
\label{eqn:qftLecture14:300}
e^{-i H T} sectionumInt \ket{n}\bra{n}
=
\sum_k \int \frac{d^3 p}{(2 \pi)^3} \ket{\Bp, k} e^{-i E_{\Bp, k}} \bra{\Bp, k}.

If we take matrix elements
\label{eqn:qftLecture14:320}
\begin{aligned}
\bra{A}
e^{-i H T} sectionumInt \ket{n}\bra{n} \ket{B}
&=
\sum_k \int \frac{d^3 p}{(2 \pi)^3} \braket{A}{\Bp, k} e^{-i E_{\Bp, k}} \braket{\Bp, k}{B} \\
&=
\sum_k \int \frac{d^3 p}{(2 \pi)^3} e^{-i E_{\Bp, k}} f(\Bp).
\end{aligned}

If we assume that $$f(\Bp)$$ is a well behaved smooth function, we have “infinite” frequency oscillation within the envelope provided by the amplitude of that function, as depicted in fig. 1.
The Riemann-Lebesgue lemma [2] describes such integrals, the result of which is that such an integral goes to zero. This is a different sort of hand waving argument, but either way, we can argue that only the ground state contributes to the sum \ref{eqn:qftLecture14:220} above.

fig. 1. High frequency oscillations within envelope of well behaved function.

### Ground state of the perturbed Hamiltonian.

With the excited states ignored, we are left with
\label{eqn:qftLecture14:340}
e^{-i H T} \ket{0} = e^{-i E_0 T} \ket{\Omega}\braket{\Omega}{0}

in the $$T \rightarrow \infty(1 – i \epsilon)$$ limit. We can now write the ground state as

\label{eqn:qftLecture14:360}
\begin{aligned}
\ket{\Omega}
&=
\evalbar{
\frac{ e^{i E_0 T – i H T } \ket{0} }{
\braket{\Omega}{0}
}
}{ T \rightarrow \infty(1 – i \epsilon) } \\
&=
\evalbar{
\frac{ e^{- i H T } \ket{0} }{
e^{-i E_0 T} \braket{\Omega}{0}
}
}{ T \rightarrow \infty(1 – i \epsilon) }.
\end{aligned}

Shifting the very large $$T \rightarrow T + t_0$$ shouldn’t change things, so
\label{eqn:qftLecture14:480}
\ket{\Omega}
=
\evalbar{
\frac{ e^{- i H (T + t_0) } \ket{0} }{
e^{-i E_0 (T + t_0) } \braket{\Omega}{0}
}
}{ T \rightarrow \infty(1 – i \epsilon) }.

A bit of manipulation shows that the operator in the numerator has the structure of a time evolution operator.

### Claim: (DIY):

\Cref{eqn:qftLecture14:80}, \ref{eqn:qftLecture14:120} may be generalized to
\label{eqn:qftLecture14:400}
U(t, t’) = e^{i H_0(t – t_0)} e^{-i H(t – t’)} e^{-i H_0(t’ – t_0)} =
T \exp{\lr{ -i \int_{t’}^t H_{\text{I,int}}(t”) dt”}}.

Observe that we recover \ref{eqn:qftLecture14:120} when $$t’ = t_0$$.  Using \ref{eqn:qftLecture14:400} we find
\label{eqn:qftLecture14:520}
\begin{aligned}
U(t_0, -T) \ket{0}
&= e^{i H_0(t_0 – t_0)} e^{-i H(t_0 + T)} e^{-i H_0(-T – t_0)} \ket{0} \\
&= e^{-i H(t_0 + T)} e^{-i H_0(-T – t_0)} \ket{0} \\
&= e^{-i H(t_0 + T)} \ket{0},
\end{aligned}

where we use the fact that $$e^{i H_0 \tau} \ket{0} = \lr{ 1 + i H_0 \tau + \cdots } \ket{0} = 1 \ket{0},$$ since $$H_0 \ket{0} = 0$$.

We are left with
\label{eqn:qftLecture14:420}
\boxed{
\ket{\Omega}
= \frac{U(t_0, -T) \ket{0} }{e^{-i E_0(t_0 – (-T))} \braket{\Omega}{0}}.
}

We are close to where we want to be. Wednesday we finish off, and then start scattering and Feynman diagrams.

# References

[1] Michael E Peskin and Daniel V Schroeder. An introduction to Quantum Field Theory. Westview, 1995.

[2] Wikipedia contributors. Riemann-lebesgue lemma — Wikipedia, the free encyclopedia, 2018. URL https://en.wikipedia.org/w/index.php?title=Riemann%E2%80%93Lebesgue_lemma&oldid=856778941. [Online; accessed 29-October-2018].

## Dirac delta function potential

November 19, 2015 phy1520 , ,

Note: there’s an error below (and in the associated PDF).  10 points to anybody that finds it (I’ve fixed it in my working version of phy1520.pdf)

### Q:Dirac delta function potential

Problem 2.24/2.25 [1] introduces a Dirac delta function potential

\label{eqn:diracPotential:20}
H = \frac{p^2}{2m} – V_0 \delta(x),

which vanishes after $$t = 0$$. Solve for the bound state for $$t < 0$$ and then the time evolution after that.

### A:

The first part of this problem was assigned back in phy356, where we solved this for a rectangular potential that had the limiting form of a delta function potential. However, this problem can be solved directly by considering the $$\Abs{x} > 0$$ and $$x = 0$$ regions separately.

For $$\Abs{x} > 0$$ Schrodinger’s equation takes the form

\label{eqn:diracPotential:40}
E \psi = -\frac{\Hbar^2}{2m} \frac{d^2 \psi}{dx^2}.

With

\label{eqn:diracPotential:60}
\kappa =
\frac{\sqrt{-2 m E}}{\Hbar},

this has solutions

\label{eqn:diracPotential:80}
\psi = e^{\pm \kappa x}.

For $$x > 0$$ we must have
\label{eqn:diracPotential:100}
\psi = a e^{-\kappa x},

and for $$x < 0$$
\label{eqn:diracPotential:120}
\psi = b e^{\kappa x}.

requiring that $$\psi$$ is continuous at $$x = 0$$ means $$a = b$$, or

\label{eqn:diracPotential:140}
\psi = \psi(0) e^{-\kappa \Abs{x}}.

For the $$x = 0$$ region, consider an interval $$[-\epsilon, \epsilon]$$ region around the origin. We must have

\label{eqn:diracPotential:160}
E \int_{-\epsilon}^\epsilon \psi(x) dx = \frac{-\Hbar^2}{2m} \int_{-\epsilon}^\epsilon \frac{d^2 \psi}{dx^2} dx – V_0 \int_{-\epsilon}^\epsilon \delta(x) \psi(x) dx.

The RHS is zero

\label{eqn:diracPotential:180}
E \int_{-\epsilon}^\epsilon \psi(x) dx
=
E \frac{ e^{-\kappa (\epsilon)} – 1}{-\kappa}
-E \frac{ 1 – e^{\kappa (-\epsilon)}}{\kappa}
\rightarrow
0.

That leaves
\label{eqn:diracPotential:200}
\begin{aligned}
V_0 \int_{-\epsilon}^\epsilon \delta(x) \psi(x) dx
&=
\frac{-\Hbar^2}{2m} \int_{-\epsilon}^\epsilon \frac{d^2 \psi}{dx^2} dx \\
&=
\frac{-\Hbar^2}{2m} \evalrange{\frac{d \psi}{dx}}{-\epsilon}{\epsilon} \\
&=
\frac{-\Hbar^2}{2m}
\psi(0)
\lr
{
-\kappa e^{-\kappa (\epsilon)}

\kappa e^{\kappa (-\epsilon)}
}.
\end{aligned}

In the $$\epsilon \rightarrow 0$$ limit this gives

\label{eqn:diracPotential:220}
V_0 = \frac{\Hbar^2 \kappa}{m}.

Equating relations for $$\kappa$$ we have

\label{eqn:diracPotential:240}
\kappa = \frac{m V_0}{\Hbar^2} = \frac{\sqrt{-2 m E}}{\Hbar},

or

\label{eqn:diracPotential:260}
E = -\inv{2 m} \lr{ \frac{m V_0}{\Hbar} }^2,

with

\label{eqn:diracPotential:280}
\psi(x, t < 0) = C \exp\lr{ -i E t/\hbar – \kappa \Abs{x}}.

The normalization requires

\label{eqn:diracPotential:300}
1
= 2 \Abs{C}^2 \int_0^\infty e^{- 2 \kappa x} dx
= 2 \Abs{C}^2 \evalrange{\frac{e^{- 2 \kappa x}}{-2 \kappa}}{0}{\infty}
= \frac{\Abs{C}^2}{\kappa},

so
\label{eqn:diracPotential:320}
\boxed{
\psi(x, t < 0) = \inv{\sqrt{\kappa}} \exp\lr{ -i E t/\hbar – \kappa \Abs{x}}. } There is only one bound state for such a potential. After turning off the potential, any plane wave $$\label{eqn:diracPotential:360} \psi(x, t) = e^{i k x – i E(k) t/\Hbar},$$ where $$\label{eqn:diracPotential:380} k = \frac{\sqrt{2 m E}}{\Hbar},$$ is a solution. In particular, at $$t = 0$$, the wave packet $$\label{eqn:diracPotential:400} \psi(x,0) = \inv{\sqrt{2\pi}} \int_{-\infty}^\infty e^{i k x} A(k) dk,$$ is a solution. To solve for $$A(k)$$, we require $$\label{eqn:diracPotential:420} \inv{\sqrt{2\pi}} \int_{-\infty}^\infty e^{i k x} A(k) dk = \inv{\sqrt{\kappa}} e^{ – \kappa \Abs{x} },$$ or $$\label{eqn:diracPotential:440} \boxed{ A(k) = \inv{\sqrt{2\pi \kappa}} \int_{-\infty}^\infty e^{-i k x} e^{ – m V_0 \Abs{x}/\Hbar^2 } dx. }$$ The initial time state established by the delta function potential evolves as \label{eqn:diracPotential:480} \boxed{ \psi(x, t > 0) = \inv{\sqrt{2\pi}} \int_{-\infty}^\infty e^{i k x – i \Hbar k^2 t/2m} A(k) dk.
}

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

## Two spin time evolution

November 14, 2015 phy1520 , , , ,

## Motivation

Our midterm posed a (low mark “quick question”) that I didn’t complete (or at least not properly). This shouldn’t have been a difficult question, but I spend way too much time on it, costing me time that I needed for other questions.

It turns out that there isn’t anything fancy required for this question, just perseverance and careful work.

## Guts

The question asked for the time evolution of a two particle state

\label{eqn:twoSpinHamiltonian:20}
\psi = \inv{\sqrt{2}} \lr{ \ket{\uparrow \downarrow} – \ket{\downarrow \uparrow} }

under the action of the Hamiltonian

\label{eqn:twoSpinHamiltonian:40}
H = – B S_{z,1} + 2 B S_{x,2} = \frac{\Hbar B}{2}\lr{ -\sigma_{z,1} + 2 \sigma_{x,2} } .

We have to know the action of the Hamiltonian on all the states

\label{eqn:twoSpinHamiltonian:60}
\begin{aligned}
H \ket{\uparrow \uparrow} &= \frac{B \Hbar}{2} \lr{ -\ket{\uparrow \uparrow} + 2 \ket{\uparrow \downarrow} } \\
H \ket{\uparrow \downarrow} &= \frac{B \Hbar}{2} \lr{ -\ket{\uparrow \downarrow} + 2 \ket{\uparrow \uparrow} } \\
H \ket{\downarrow \uparrow} &= \frac{B \Hbar}{2} \lr{ \ket{\downarrow \uparrow} + 2 \ket{\downarrow \downarrow} } \\
H \ket{\downarrow \downarrow} &= \frac{B \Hbar}{2} \lr{ \ket{\downarrow \downarrow} + 2 \ket{\downarrow \uparrow} } \\
\end{aligned}

With respect to the basis $$\setlr{ \ket{\uparrow \uparrow}, \ket{\uparrow \downarrow}, \ket{\downarrow \uparrow}, \ket{\downarrow \downarrow} }$$, the matrix of the Hamiltonian is

\label{eqn:twoSpinHamiltonian:80}
H =
\frac{ \Hbar B }{2}
\begin{bmatrix}
-1 & 2 & 0 & 0 \\
2 & -1 & 0 & 0 \\
0 & 0 & 1 & 2 \\
0 & 0 & 2 & 1 \\
\end{bmatrix}

Utilizing the block diagonal form (and ignoring the $$\Hbar B/2$$ factor for now), the characteristic equation is

\label{eqn:twoSpinHamiltonian:100}
0
=
\begin{vmatrix}
-1 -\lambda & 2 \\
2 & -1 – \lambda
\end{vmatrix}
\begin{vmatrix}
1 -\lambda & 2 \\
2 & 1 – \lambda
\end{vmatrix}
=
\lr{(1 + \lambda)^2 – 4}
\lr{(1 – \lambda)^2 – 4}.

This has solutions

\label{eqn:twoSpinHamiltonian:120}
1 \pm \lambda = \pm 2,

or, with the $$\Hbar B/2$$ factors put back in

\label{eqn:twoSpinHamiltonian:140}
\lambda = \pm \Hbar B/2 , \pm 3 \Hbar B/2.

I was thinking that we needed to compute the time evolution operator

\label{eqn:twoSpinHamiltonian:160}
U = e^{-i H t/\Hbar},

but we actually only need the eigenvectors, and the inverse relations. We can find the eigenvectors by inspection in each case from

\label{eqn:twoSpinHamiltonian:180}
\begin{aligned}
H – (1) \frac{ \Hbar B }{2}
&=
\frac{ \Hbar B }{2}
\begin{bmatrix}
-2 & 2 & 0 & 0 \\
2 & -2 & 0 & 0 \\
0 & 0 & 0 & 2 \\
0 & 0 & 2 & 0 \\
\end{bmatrix} \\
H – (-1) \frac{ \Hbar B }{2}
&=
\frac{ \Hbar B }{2}
\begin{bmatrix}
0 & 2 & 0 & 0 \\
2 & 0 & 0 & 0 \\
0 & 0 & 2 & 2 \\
0 & 0 & 2 & 2 \\
\end{bmatrix} \\
H – (3) \frac{ \Hbar B }{2}
&=
\frac{ \Hbar B }{2}
\begin{bmatrix}
-4 & 2 & 0 & 0 \\
2 & -4 & 0 & 0 \\
0 & 0 &-2 & 2 \\
0 & 0 & 2 &-2 \\
\end{bmatrix} \\
H – (-3) \frac{ \Hbar B }{2}
&=
\frac{ \Hbar B }{2}
\begin{bmatrix}
2 & 2 & 0 & 0 \\
2 & 2 & 0 & 0 \\
0 & 0 & 4 & 2 \\
0 & 0 & 2 & 1 \\
\end{bmatrix}.
\end{aligned}

The eigenkets are

\label{eqn:twoSpinHamiltonian:280}
\begin{aligned}
\ket{1} &=
\inv{\sqrt{2}}
\begin{bmatrix}
1 \\
1 \\
0 \\
0 \\
\end{bmatrix} \\
\ket{-1} &=
\inv{\sqrt{2}}
\begin{bmatrix}
0 \\
0 \\
1 \\
-1 \\
\end{bmatrix} \\
\ket{3} &=
\inv{\sqrt{2}}
\begin{bmatrix}
0 \\
0 \\
1 \\
1 \\
\end{bmatrix} \\
\ket{-3} &=
\inv{\sqrt{2}}
\begin{bmatrix}
1 \\
-1 \\
0 \\
0 \\
\end{bmatrix},
\end{aligned}

or

\label{eqn:twoSpinHamiltonian:300}
\begin{aligned}
\sqrt{2} \ket{1} &= \ket{\uparrow \uparrow} + \ket{\uparrow \downarrow} \\
\sqrt{2} \ket{-1} &= \ket{\downarrow \uparrow} – \ket{\downarrow \downarrow} \\
\sqrt{2} \ket{3} &= \ket{\downarrow \uparrow} + \ket{\downarrow \downarrow} \\
\sqrt{2} \ket{-3} &= \ket{\uparrow \uparrow} – \ket{\uparrow \downarrow}.
\end{aligned}

We can invert these

\label{eqn:twoSpinHamiltonian:220}
\begin{aligned}
\ket{\uparrow \uparrow} &= \inv{\sqrt{2}} \lr{ \ket{1} + \ket{-3} } \\
\ket{\uparrow \downarrow} &= \inv{\sqrt{2}} \lr{ \ket{1} – \ket{-3} } \\
\ket{\downarrow \uparrow} &= \inv{\sqrt{2}} \lr{ \ket{3} + \ket{-1} } \\
\ket{\downarrow \downarrow} &= \inv{\sqrt{2}} \lr{ \ket{3} – \ket{-1} } \\
\end{aligned}

The original state of interest can now be expressed in terms of the eigenkets

\label{eqn:twoSpinHamiltonian:240}
\psi
=
\inv{2} \lr{
\ket{1} – \ket{-3} –
\ket{3} – \ket{-1}
}

The time evolution of this ket is

\label{eqn:twoSpinHamiltonian:260}
\begin{aligned}
\psi(t)
&=
\inv{2}
\lr{
e^{-i B t/2} \ket{1}
– e^{3 i B t/2} \ket{-3}
– e^{-3 i B t/2} \ket{3}
– e^{i B t/2} \ket{-1}
} \\
&=
\inv{2 \sqrt{2}}
\Biglr{
e^{-i B t/2} \lr{ \ket{\uparrow \uparrow} + \ket{\uparrow \downarrow} }
– e^{3 i B t/2} \lr{ \ket{\uparrow \uparrow} – \ket{\uparrow \downarrow} }
– e^{-3 i B t/2} \lr{ \ket{\downarrow \uparrow} + \ket{\downarrow \downarrow} }
– e^{i B t/2} \lr{ \ket{\downarrow \uparrow} – \ket{\downarrow \downarrow} }
} \\
&=
\inv{2 \sqrt{2}}
\Biglr{
\lr{ e^{-i B t/2} – e^{3 i B t/2} } \ket{\uparrow \uparrow}
+ \lr{ e^{-i B t/2} + e^{3 i B t/2} } \ket{\uparrow \downarrow}
– \lr{ e^{-3 i B t/2} + e^{i B t/2} } \ket{\downarrow \uparrow}
+ \lr{ e^{i B t/2} – e^{-3 i B t/2} } \ket{\downarrow \downarrow}
} \\
&=
\inv{2 \sqrt{2}}
\Biglr{
e^{i B t/2} \lr{ e^{-2 i B t/2} – e^{2 i B t/2} } \ket{\uparrow \uparrow}
+ e^{i B t/2} \lr{ e^{-2 i B t/2} + e^{2 i B t/2} } \ket{\uparrow \downarrow}
– e^{- i B t/2} \lr{ e^{-2 i B t/2} + e^{2 i B t/2} } \ket{\downarrow \uparrow}
+ e^{- i B t/2} \lr{ e^{2 i B t/2} – e^{-2 i B t/2} } \ket{\downarrow \downarrow}
} \\
&=
\inv{\sqrt{2}}
\lr{
i \sin( B t )
\lr{
e^{- i B t/2} \ket{\downarrow \downarrow} – e^{i B t/2} \ket{\uparrow \uparrow}
}
+ \cos( B t ) \lr{
e^{i B t/2} \ket{\uparrow \downarrow}
– e^{- i B t/2} \ket{\downarrow \uparrow}
}
}
\end{aligned}

Note that this returns to the original state when $$t = \frac{2 \pi n}{B}, n \in \mathbb{Z}$$. I think I’ve got it right this time (although I got a slightly different answer on paper before typing it up.)

This doesn’t exactly seem like a quick answer question, at least to me. Is there some easier way to do it?

## Second update of aggregate notes for phy1520, Graduate Quantum Mechanics

I’ve posted a second update of my aggregate notes for PHY1520H Graduate Quantum Mechanics, taught by Prof. Arun Paramekanti. In addition to what was noted previously, this contains lecture notes up to lecture 9, my ungraded solutions for the second problem set, and some additional worked practise problems.

Most of the content was posted individually in the following locations, but those original documents will not be maintained individually any further.

## Time evolution of spin half probability and dispersion

October 15, 2015 phy1520 , , ,

## Question: Time evolution of spin half probability and dispersion ([1] pr. 2.3)

A spin $$1/2$$ system $$\BS \cdot \ncap$$, with $$\ncap = \sin \beta \xcap + \cos\beta \zcap$$, is in state with eigenvalue $$\Hbar/2$$, acted on by a magnetic field of strength $$B$$ in the $$+z$$ direction.

## (a)

If $$S_x$$ is measured at time $$t$$, what is the probability of getting $$+ \Hbar/2$$?

## (b)

Evaluate the dispersion in $$S_x$$ as a function of t, that is,

\label{eqn:spinTimeEvolution:20}
\expectation{\lr{ S_x – \expectation{S_x}}^2}.

## (c)

Check your answers for $$\beta \rightarrow 0, \pi/2$$ to see if they make sense.

## (a)

The spin operator in matrix form is
\label{eqn:spinTimeEvolution:40}
\begin{aligned}
S \cdot \ncap
&=
\frac{\Hbar}{2} \lr{ \sigma_z \cos\beta + \sigma_x \sin\beta } \\
&=
\frac{\Hbar}{2} \lr{ \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} \cos\beta + \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \sin\beta } \\
&=
\frac{\Hbar}{2}
\begin{bmatrix}
\cos\beta & \sin\beta \\
\sin\beta & -\cos\beta
\end{bmatrix}.
\end{aligned}

The $$\ket{S \cdot \ncap ; + }$$ eigenstate is found from

\label{eqn:spinTimeEvolution:60}
\lr{ S \cdot \ncap – \Hbar/2}
\begin{bmatrix}
a \\
b
\end{bmatrix}
= 0,

or

\label{eqn:spinTimeEvolution:80}
\begin{aligned}
0
&=
\lr{ \cos\beta – 1 } a + \sin\beta b \\
&=
\lr{ -2 \sin^2(\beta/2) } a + 2 \sin(\beta/2) \cos(\beta/2) b \\
&=
\lr{ – \sin(\beta/2) } a + \cos(\beta/2) b,
\end{aligned}

or

\label{eqn:spinTimeEvolution:100}
\ket{ S \cdot \ncap ; + }
=
\begin{bmatrix}
\cos(\beta/2) \\
\sin(\beta/2) \\
\end{bmatrix}.

The Hamiltonian is

\label{eqn:spinTimeEvolution:120}
H
= – \frac{e B}{m c} S_z
= – \frac{e B \Hbar}{2 m c} \sigma_z,

so the time evolution operator is

\label{eqn:spinTimeEvolution:140}
U
= e^{-i H t/\Hbar}
= e^{ \frac{i e B t }{2 m c} \sigma_z }.

Let $$\omega = e B/(2 m c)$$, so

\label{eqn:spinTimeEvolution:160}
\begin{aligned}
U
&=
e^{i \sigma_z \omega t} \\
&=
\cos(\omega t) + i \sigma_z \sin(\omega t) \\
&=
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
\cos(\omega t)
+
i \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} \sin(\omega t) \\
&=
\begin{bmatrix}
e^{i \omega t} & 0 \\
0 & e^{-i \omega t}
\end{bmatrix}.
\end{aligned}

The time evolution of the initial state is

\label{eqn:spinTimeEvolution:180}
\begin{aligned}
\ket{S \cdot \ncap ; + }(t)
&=
U \ket{S \cdot \ncap ; + }(0) \\
&=
\begin{bmatrix}
e^{i \omega t} & 0 \\
0 & e^{-i \omega t}
\end{bmatrix}
\begin{bmatrix}
\cos(\beta/2) \\
\sin(\beta/2) \\
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos(\beta/2) e^{i \omega t} \\
\sin(\beta/2) e^{-i \omega t} \\
\end{bmatrix}.
\end{aligned}

The probability of finding the state in $$\ket{S \cdot \xcap ; + }$$ at time $$t$$ (i.e. measuring $$S_x$$ and finding $$\Hbar/2$$) is

\label{eqn:spinTimeEvolution:200}
\begin{aligned}
\Abs{\braket{S \cdot \xcap ; + }{S \cdot \ncap ; + }}^2
&=
\Abs{\inv{\sqrt{2}}
\begin{bmatrix}
1 & 1 \\
\end{bmatrix}
\begin{bmatrix}
\cos(\beta/2) e^{i \omega t} \\
\sin(\beta/2) e^{-i \omega t} \\
\end{bmatrix}
}^2 \\
&=
\inv{2}
\Abs{
\cos(\beta/2) e^{i \omega t} +
\sin(\beta/2) e^{-i \omega t} }^2 \\
&=
\inv{2} \lr{ 1 + 2 \cos(\beta/2) \sin(\beta/2) \cos(2 \omega t) } \\
&=
\inv{2} \lr{ 1 + \sin(\beta) \cos( 2 \omega t) }.
\end{aligned}

## (b)

To calculate the dispersion first note that

\label{eqn:spinTimeEvolution:300}
S_x^2
= \lr{ \frac{\Hbar}{2} }^2 \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}^2
= \lr{ \frac{\Hbar}{2} }^2,

so only the first order expectation is non-trivial to calculate. That is

\label{eqn:spinTimeEvolution:320}
\begin{aligned}
\expectation{S_x}
&=
\frac{\Hbar}{2}
\begin{bmatrix}
\cos(\beta/2) e^{-i \omega t} &
\sin(\beta/2) e^{i \omega t}
\end{bmatrix}
\begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}
\begin{bmatrix}
\cos(\beta/2) e^{i \omega t} \\
\sin(\beta/2) e^{-i \omega t} \\
\end{bmatrix} \\
&=
\frac{\Hbar}{2}
\begin{bmatrix}
\cos(\beta/2) e^{-i \omega t} &
\sin(\beta/2) e^{i \omega t}
\end{bmatrix}
\begin{bmatrix}
\sin(\beta/2) e^{-i \omega t} \\
\cos(\beta/2) e^{i \omega t} \\
\end{bmatrix} \\
&=
\frac{\Hbar}{2}
\sin(\beta/2) \cos(\beta/2) \lr{ e^{-2 i \omega t} + e^{ 2 i \omega t} } \\
&=
\frac{\Hbar}{2} \sin\beta \cos( 2 \omega t ).
\end{aligned}

This gives

\label{eqn:spinTimeEvolution:340}
\boxed{
\expectation{(\Delta S_x)^2}
=
\lr{ \frac{\Hbar}{2} }^2 \lr{ 1 – \sin^2\beta \cos^2( 2 \omega t ) }.
}

## (c)

For $$\beta = 0$$, $$\ncap = \zcap$$, and $$\beta = \pi/2$$, $$\ncap = \xcap$$. For the first case, the state is in an eigenstate of $$S_z$$, so must evolve as

\label{eqn:spinTimeEvolution:220}
\ket{S \cdot \ncap ; + }(t) = \ket{S \cdot \ncap ; + }(0) e^{i \omega t}.

The probability of finding it in state $$\ket{S \cdot \xcap ; + }$$ is therefore

\label{eqn:spinTimeEvolution:240}
\begin{aligned}
\Abs{
\inv{\sqrt{2}}
\begin{bmatrix}
1 & 1
\end{bmatrix}
\begin{bmatrix}
e^{i \omega t} \\
0
\end{bmatrix}
}^2
&=
\inv{2} \Abs{ e^{i\omega t} }^2 \\
&=
\inv{2} \\
&=
\inv{2} \lr{ 1 + \sin(0) \cos(2 \omega t) }.
\end{aligned}

This matches \ref{eqn:spinTimeEvolution:200} as expected.

For $$\beta = \pi/2$$ we have

\label{eqn:spinTimeEvolution:260}
\begin{aligned}
\ket{S \cdot \xcap ; + }(t)
&=
\inv{\sqrt{2}}
\begin{bmatrix}
e^{i \omega t} & 0 \\
0 & e^{-i \omega t}
\end{bmatrix}
\begin{bmatrix}
1 \\
1
\end{bmatrix} \\
&=
\inv{\sqrt{2}}
\begin{bmatrix}
e^{i \omega t} \\
e^{-i \omega t}
\end{bmatrix}.
\end{aligned}

The probability for the $$\Hbar/2$$ $$S_x$$ measurement at time $$t$$ is
\label{eqn:spinTimeEvolution:280}
\begin{aligned}
\Abs{
\inv{2}
\begin{bmatrix}
1 & 1
\end{bmatrix}
\begin{bmatrix}
e^{i \omega t} \\
e^{-i \omega t}
\end{bmatrix}
}^2
&=
\inv{4} \Abs{ e^{i \omega t} + e^{-i \omega t} }^2 \\
&=
\cos^2(\omega t) \\
&=
\inv{2}\lr{ 1 + \sin(\pi/2) \cos( 2 \omega t )}.
\end{aligned}

Again, this matches the expected value.

For the dispersions, at $$\beta = 0$$, the dispersion is

\label{eqn:spinTimeEvolution:360}
\lr{\frac{\Hbar}{2}}^2

This is the maximum dispersion, which makes sense since we are measuring $$S_x$$ when the initial state is $$\ket{S \cdot \zcap ; + }$$. For $$\beta = \pi/2$$ the dispersion is

\label{eqn:spinTimeEvolution:380}
\lr{\frac{\Hbar}{2}}^2 \sin^2 ( 2 \omega t ).

This starts off as zero dispersion (because the initial state is $$\ket{ S \cdot \xcap ; + }$$, but then oscillates.

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.